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Use multiplication instead of division by a constant
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{-# LANGUAGE BangPatterns #-} | ||
{-# LANGUAGE MagicHash #-} | ||
{-# LANGUAGE UnboxedTuples #-} | ||
-- | | ||
-- Module : Data.Unicode.Internal.Division | ||
-- Copyright : (c) 2020 Andrew Lelechenko | ||
-- | ||
-- License : BSD-3-Clause | ||
-- Maintainer : harendra.kumar@gmail.com | ||
-- Stability : experimental | ||
-- Portability : GHC | ||
-- | ||
-- Fast division by known constants. | ||
-- | ||
-- Division by a constant can be replaced by a double-word multiplication. | ||
-- Roughly speaking, instead of dividing by x, multiply by 2^64/x, | ||
-- obtaining 128-bit-long product, and take upper 64 bits. The peculiar | ||
-- details can be found in Hacker's Delight, Ch. 10. | ||
-- | ||
-- Even GHC 8.10 does not provide a primitive for a signed double-word | ||
-- multiplication, but since our applications does not involve negative | ||
-- integers, we convert 'Int' to 'Word' and use 'GHC.Exts.timesWord#'. | ||
-- | ||
-- Textbook unsigned division by 21 or 28 becomes involved, when an argument | ||
-- is allowed to take the full range of 'Word' up to 2^64. Luckily, in our | ||
-- case the argument was casted from 'Int', so we can guarantee that it is | ||
-- below 2^63. | ||
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module Data.Unicode.Internal.Division | ||
( quotRem21 | ||
, quotRem28 | ||
) where | ||
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import Data.Bits (Bits(..), FiniteBits(..)) | ||
import GHC.Exts (Word(..), timesWord2#) | ||
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highMul :: Word -> Word -> Word | ||
highMul (W# x#) (W# y#) = W# high# | ||
where | ||
!(# high#, _ #) = timesWord2# x# y# | ||
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-- Input must be non-negative. | ||
-- | ||
-- Instead of division by 21, we compute | ||
-- floor(floor((2^68+17)/21 * n) / 2^68) = floor((2^68+17)/21 * n/2^68) = | ||
-- floor(n/21 + (n/2^63 * 17/32)/21) = floor(n/21), | ||
-- because n/2^63 * 17/32 < 1. | ||
{-# INLINE quotRem21 #-} | ||
quotRem21 :: Int -> (Int, Int) | ||
quotRem21 n | ||
| finiteBitSize (0 :: Word) /= 64 | ||
= n `quotRem` 21 | ||
| otherwise | ||
= (fromIntegral q, fromIntegral (w - 21 * q)) | ||
where | ||
w = fromIntegral n | ||
high = highMul w 14054662151397753613 -- (2^68+17)/21 | ||
q = high `shiftR` 4 | ||
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-- Input must be non-negative. | ||
-- | ||
-- Instead of division by 28, we compute | ||
-- floor(floor((2^65+3)/7 * n) / 2^67) = floor((2^65+3)/7 * n/2^67) = | ||
-- floor(n/28 + (n/2^63 * 3/4)/28) = floor(n/28), | ||
-- because n/2^63 * 3/4 < 1. | ||
{-# INLINE quotRem28 #-} | ||
quotRem28 :: Int -> (Int, Int) | ||
quotRem28 n | ||
| finiteBitSize (0 :: Word) /= 64 | ||
= n `quotRem` 28 | ||
| otherwise | ||
= (fromIntegral q, fromIntegral r) | ||
where | ||
w = fromIntegral n | ||
high = highMul w 5270498306774157605 -- (2^65+3)/7 | ||
q = high `shiftR` 3 | ||
prod = (q `shiftL` 3 - q) `shiftL` 2 | ||
r = w - prod |
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